Difference between revisions of "HW3.B Allen Humphreys ECE301Fall2008mboutin" - Rhea

Latest revision as of 13:29, 19 September 2008

We know that the system is linear, therefore, we can sum the inputs to equal the sum of outputs:

The input is

x_{c}(t) = e^{2\times jt} + e^{-2\times jt}

and the corresponding output is

y_{c}(t) = t \times e^{-2jt} + t \times e^{2jt}

We know by Euler's formula:

\cos 2t = {e^{2jt} + e^{-2jt} \over 2}

Finally, by the multiplication property of linear systems:

The input,

{x_{c}(t) \over 2} = \cos 2t = {e^{2\times jt} + e^{-2\times jt} \over 2}

will yield the output:

{y_{c}(t) \over 2} = {t e^{-2jt} + t e^{2jt} \over 2}

@@ Line 1: / Line 1: @@
 [[Homework 3_ECE301Fall2008mboutin]] - [[HW3.A Allen Humphreys_ECE301Fall2008mboutin|'''A''']] - [[HW3.B Allen Humphreys_ECE301Fall2008mboutin|'''B''']] - [[HW3.C Allen Humphreys_ECE301Fall2008mboutin|'''C''']]
-We know that the system is linear, therefore, we can conclude the following about the given information.
+We know that the system is linear, therefore, we can sum the inputs to equal the sum of outputs:
-<math>x_{1}(t) = e^{2\times jt}</math> and <math>x_{2}(t) = e^{-2\times jt}</math>
+The input is
+: <math>x_{c}(t) = e^{2\times jt} + e^{-2\times jt} </math>
+and the corresponding output is
+:<math>y_{c}(t) = t \times e^{-2jt} + t \times e^{2jt}</math>
-<math>y_{1}(t) = t \times e^{-2jt}</math> and <math>y_{2}(t) = t \times e^{2jt}</math>
+We know by Euler's formula:
-By linearity,
+: <math>\cos 2t = {e^{2jt} + e^{-2jt} \over 2}</math>
+Finally, by the multiplication property of linear systems:
+The input,
+:<math>   {x_{c}(t) \over 2} = \cos 2t = {e^{2\times jt} + e^{-2\times jt} \over 2} </math>
+will yield the output:
+:<math>{y_{c}(t) \over 2} = {t e^{-2jt} + t e^{2jt} \over 2}</math>